Section 8.5

Solid Mechanics Part I Kelly 260

8.5 Virtual Work Consider a mass attached to a spring and pulled by an applied force aplF , Fig. 8.5.1a.

When the mass is in equilibrium, 0 aplspr FF , where kxFspr is the spring force

with x the distance from the spring reference position.

Figure 8.5.1: a force extending an elastic spring; (a) block in equilibrium, (b) block

not at its equilibrium position In order to develop a number of powerful techniques based on a concept known as virtual work, imagine that the mass is not in fact at its equilibrium position but at an (incorrect) non-equilibrium position xx , Fig. 8.5.1b. The imaginary displacement x is called a virtual displacement. Define the virtual work W done by a force to be the equilibrium force times this small imaginary displacement x . It should be emphasized that virtual work is not real work – no work has been performed since x is not a real displacement which has taken place; this is more like a “thought experiment”. The virtual work of the spring force is then xkxxFW sprspr . The virtual work of the applied

force is xFW aplapl . The total virtual work is

xFkxWWW aplaplspr (8.5.1)

There are two ways of viewing this expression. First, if the system is in equilibrium ( 0 aplFkx ) then the virtual work is zero, 0W . Alternatively, if the virtual work

is zero then, since x is arbitrary, the system must be in equilibrium. Thus the virtual work idea gives one an alternative means of determining whether a system is in equilibrium. The symbol is called a variation so that, for example, x is a variation in the displacement (from equilibrium). Virtual work is explored further in the following section.

sprF

aplF

sprF

aplF

ex

x

)a(

)b(

Section 8.5

Solid Mechanics Part I Kelly 261

8.5.1 Principle of Virtual Work: a single particle A particle of mass m is acted upon by a number of forces, Nfff ,,, 21 , Fig. 8.5.2.

Suppose the particle undergoes a virtual displacement u ; to reiterate, these impressed forces if do not cause the particle to move, one imagines it to be incorrectly positioned a

little away from the true equilibrium position.

Figure 8.5.2: a particle in equilibrium under the action of a number of forces If the particle is moving with an acceleration a, the quantity am is treated as an inertial force. The total virtual work is then (each term here is the dot product of two vectors)

uaf

mWN

ii

1

(8.5.2)

Now if the particle is in equilibrium by the action of the effective (impressed plus inertial) force, then

0W (8.5.3) This can be expressed as follows: The principle of virtual work (or principle of virtual displacements) I: if a particle is in equilibrium under the action of a number of forces (including the inertial force) the total work done by the forces for a virtual displacement is zero Alternatively, one can define the external virtual work uf iWext and the virtual

kinetic energy ua mK in which case the principle takes the form KW ext

(compare with the work-energy principle, Eqn. 8.1.10). In the above, the principle of virtual work was derived using Newton’s second law. One could just as well regard the principle of virtual work as the fundamental principle and from it derive the conditions for equilibrium. In this case one can say that1

1 note the word any here: this must hold for all possible virtual displacements, for it will always be possible

to find one virtual displacement which is perpendicular to the resultant of the forces, so that 0 uf

even though f is not necessarily zero

u1f

2f 3f

Section 8.5

Solid Mechanics Part I Kelly 262

The principle of virtual work (or principle of virtual displacements) II: a particle is in equilibrium under the action of a system of forces (including the inertial force) if the total work done by the forces is zero for any virtual displacement of the particle Constraints In many practical problems, the particle will usually be constrained to move in only certain directions. For example consider a ball rolling over a table, Fig. 8.5.3. If the ball is in equilibrium then all the forces sum to zero, m R f a 0 , where one

distinguishes between the non-reaction forces if and the reaction force R. If the virtual

displacement u is such that the constraint is not violated, that is the ball is not allowed to go “through” the table, then u and R are perpendicular, the virtual work done by the reaction force is zero and 0 uaf mW . This is one of the benefits of the

principle of virtual work; one does not need to calculate the forces of constraint R in order to determine the forces if which maintain the particle in equilibrium.

Figure 8.5.3: a particle constrained to move over a surface The term kinematically admissible displacement is used to mean one that does not violate the constraints, and hence one arrives at the version of the principle which is often used in practice: The principle of virtual work (or principle of virtual displacements) III: a particle is in equilibrium under the action of a system of forces (including the inertial force) if the total work done by the forces (excluding reaction forces) is zero for any kinematically admissible virtual displacement of the particle Whether one uses a kinematically admissible virtual displacement and so disregard reaction forces, or permit a virtual displacement that violates the constraint conditions will usually depend on the problem at hand. In this next example, use is made of a kinematically inadmissible virtual displacement. Example Consider a rigid bar of length L supported at its ends and loaded by a force F a distance a from the left hand end, Fig. 8.5.3a. Reaction forces CA RR , act at the ends. Let point C

undergo a virtual displacement u . From similar triangles, the displacement at B is uLa )/( . End A does not move and so no virtual work is performed there. The total

virtual work is

R

1f2f

3f

Section 8.5

Solid Mechanics Part I Kelly 263

uL

aFuRW C (8.5.4)

Note the minus sign here – the displacement at B is in a direction opposite to that of the action of the load and hence the work is negative. The beam is in equilibrium when

0W and hence LaFRC / .

Figure 8.5.3: a loaded rigid bar; (a) bar geometry, (b) a virtual displacement at end

C ■

8.5.2 Principle of Virtual Work: deformable bodies A deformable body can be imagined to undergo virtual displacements (not necessarily the same throughout the body). Virtual work is done by the externally applied forces – external virtual work – and by the internal forces – internal virtual work. Looking again at the spring problem of Fig. 8.5.1, the external virtual work is xFW aplext and,

considering the spring force to be an “internal” force, the internal virtual work is xkxW int . This latter virtual work can be re-written as UW int where U is

the virtual potential energy change which occurs when the spring is moved a distance x (keeping the spring force constant). In the same way, the internal virtual work of an elastic body is the (negative of the) virtual strain energy and the principle of virtual work can be expressed as

UWext Principle of Virtual Work for an Elastic Body (8.5.4)

The principle can be extended to accommodate dissipation (see Book IV), but only elastic materials will be examined here. The virtual strain energy for a uniaxial rod is derived next. 8.5.3 Virtual Strain Energy for a Uniaxially Loaded Bar In what follows, to distinguish between the strain energy and the displacement, the former will now be denoted by w and the latter by u.

A B CF

L

a

CRAR

u

CR

F

Section 8.5

Solid Mechanics Part I Kelly 264

Consider a uniaxial bar which undergoes strains . The strain is the unit change in length and, considering an element of length dx , Fig. 8.5.4a, the strain is

dx

du

x

xxuxxux

)()( (8.5.5)

in the limit as 0x . With ddw , the strain energy density is

22

2

1

2

1

2

1

dx

duEEw (8.5.6)

and the strain energy is

dxdx

duEAdV

dx

duEU

L

v

0

22

22

1 (8.5.7)

This is the actual strain energy change when the bar undergoes actual strains . For the simple case of constant A and L and constant strain Ldxdu // where is the elongation of the bar, Eqn. 8.5.7 reduces to LAEU 2/2 (equivalent to Eqn. 8.2.2).

Figure 8.5.4: element undergoing actual and virtual displacements; (a) actual displacements, (b) virtual displacements

It will now be shown that the internal virtual work done as material particles undergo virtual displacements u is given by U , with U given by Eqn. 8.5.7. Consider an element to “undergo” virtual displacements u , Fig. 8.5.4b, which are, by definition, measured from the actual displacements. The virtual displacements give rise to virtual strains:

dx

ud

x

xuxxu

)()( (8.5.8)

again in the limit as 0x . Since dxdu / , it follows that

x

)(xu )( xxu

x xx

)(xu )( xxu

(a)

(b)

Section 8.5

Solid Mechanics Part I Kelly 265

dx

ud

dx

du

(8.5.9)

In other words, the variation of the derivative is equal to the derivative of the variation2. One other result is needed before calculating the internal virtual work. Consider a function of the displacement, )(uf . The variation of f when u undergoes a virtual displacement is by definition

udu

dfu

u

ufuufufuuff

)()()()( (8.5.10)

now in the limit as the virtual displacement 0u . From this one can write

dx

du

dx

du

dx

du 22

(8.5.11)

The stress applied to the surface of the element under consideration is an “external force”. The internal force is the equal and opposite stress on the other side of the surface inside the element. The internal virtual work (per unit volume) is then W . Since is the actual stress, unaffected by the virtual straining,

22

2

1

2

1

dx

duE

dx

duE

dx

du

dx

duEEW (8.5.12)

since the Young’s modulus is unaffected by any virtual displacement. The total work done is then

dVdx

duEW

v

2

int 2

1 (8.5.13)

which, comparing with Eqn. 8.5.7, is the desired result, UW int .

Example Two rods with cross sectional areas 21 , AA , lengths 21 , LL and Young’s moduli 21 , EE and joined together with the other ends fixed, as shown in Fig. 8.5.5. The rods are subjected to a force P where they meet. As the rods elongate/contract, the strain is simply

LuB / , where Bu is the displacement of the point at which the force is applied. The total elastic strain energy is, from Eqn. 8.5.7,

2 this holds in general for any function; manipulations with variations form a part of a branch of mathematics known as the Calculus of Variations, which is concerned in the main with minima/maxima problems

Section 8.5

Solid Mechanics Part I Kelly 266

2

2

222

1

11

22 BB uL

AEu

L

AEU (8.5.14)

Introduce now a virtual displacement Bu at B. The external virtual work is

BuPW ext . The principle of virtual work, Eqn. 8.5.4, states that

2

2

22

1

11

22 BB uL

AE

L

AEuP (8.5.15)

Figure 8.8.5: two rods subjected to a force P From relation 8.5.10,

BBB uuL

AE

L

AEuP

2

22

1

11 (8.5.16)

The virtual displacement Bu is arbitrary and so can be cancelled out, giving the result

1

2

22

1

11

L

AE

L

AEPuB (8.5.17)

from which the strains and hence stresses can be evaluated. Note that the reaction forces were not involved in this solution method.

■ 8.5.4 Virtual Strain Energy for a Beam The strain energy in a beam is given by Eqn. 8.2.7, viz.

dxEI

MU

L

0

2

2 (8.5.18)

Using the moment-curvature relation 7.4.37, 22 / dxvdEIM , where v is the deflection of the beam,

P

applied force

fixed fixed 1L 2L

B

Section 8.5

Solid Mechanics Part I Kelly 267

dxdx

vdEIU

L

0

2

2

2

2 (8.5.19)

and the virtual strain energy is

dxdx

vdEIU

L

0

2

2

2

2 (8.5.20)

It is not easy to analyse problems using this expression and the principle of virtual work directly, but this expression will be used in the next section in conjunction with the related principle of minimum potential energy. 8.5.5 Problems 1. Consider a uniaxial bar of length L with constant cross section A and Young’s

modulus E, fixed at one end and subjected to a force P at the other. Use the principle of virtual work to show that the displacement at the loaded end is EAPLu / .

2. Consider a uniaxial bar of length L, cross sectional area A and Young’s modulus E.

What factor of EAL is the strain energy when the displacements in the bar are xu 310 , with x measured from one end of the bar? What is the internal virtual

work for a virtual displacement xu 510 ? For a constant virtual displacement along the bar?

3. A rigid bar rests upon three columns, a central column with Young’s modulus

GPa100 and two equidistant outer columns with Young’s moduli GPa200 . The columns are of equal length 1m and cross-sectional area 2cm1 . The rigid bar is subjected to a downward force of kN10 . Use the principle of virtual work to evaluate the vertical displacement downward of the rigid bar.

4. Re-solve problem 3 from §8.2.6 using the principle of virtual displacements.